Trigonometry!!!!!!!!!!!!!!!!!!

Question

Answer 1

The given that  2 sin²x +3sinx  = -1   for 0 ≤ x < 2π

2 sin²x +3sinx  +1 = 0

2 sin²x + 2sinx + sinx +1 = 0

2sinx ( sinx+1)+1( sinx+1) = 0

( 2sinx + 1) ( sinx +1) = 0

2sinx + 1 = 0         sinx = -1

sinx = - 1/2             x  = sin^-1( -1 )

x = sin^-1( -1/2)

Therefore x =   and  x = 7π/6  and x = 3π/2

b) csc²x - 2 = 0  for 0 ≤ x < 2π

csc²x = 2

csc x = ±√2

x = csc^-1( (±√2 )

Therefore x = π/4 and x = 5π/4.

c)sin²x +cosx = -1   for 0 ≤ x < 2π

sin²x +cosx +1 = 0      [ sin²x = 1- cos²x ]

( 1- cos²x ) + cosx +1 = 0

cos²x - cosx - 2 = 0

cos²x - 2cosx +cosx - 2 = 0

cosx ( cosx - 2) +1( cosx - 2) = 0

( cosx +1) (cosx -2) = 0

cosx = -1        cosx = 2

x = cos^-1( -1)   x = cos^-1( 2)

Therefore x = π .